An element with molar mass 2.7 times 10^{-2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given:Density, $d = 2.7 	imes 10^{3} \,kg \,m^{-3}$Molar mass, $M = 2.7 	imes 10^{-2} \,kg \,mol^{-1}$Edge length, $a = 405 \,pm = 4.05 	imes 1

Vedclass · Accepted Answer

Face-centred cubic $(fcc)$

Vedclass · Answer

(A) Given:Density, $d = 2.7 	imes 10^{3} \,kg \,m^{-3}$Molar mass, $M = 2.7 	imes 10^{-2} \,kg \,mol^{-1}$Edge length, $a = 405 \,pm = 4.05 	imes 10^{-10} \,m$Avogadro's number, $N_A = 6.022 	imes 10^{23} \,mol^{-1}$Using the formula for density of a unit cell:$d = \frac{z \cdot M}{a^3 \cdot N_A}$Rearranging for $z$:$z = \frac{d \cdot a^3 \cdot N_A}{M}$Substituting the values:$z = \frac{(2.7 	imes 10^{3}) \cdot (4.05 	imes 10^{-10})^3 \cdot (6.022 	imes 10^{23})}{2.7 	imes 10^{-2}}$$z = \frac{2.7 	imes 10^{3} \cdot 66.43 	imes 10^{-30} \cdot 6.022 	imes 10^{23}}{2.7 	imes 10^{-2}}$$z \approx 4$Since the number of atoms per unit cell is $4$, the unit cell is face-centred cubic $(fcc)$.

An element with molar mass $2.7 \times 10^{-2} \,kg \,mol^{-1}$ forms a cubic unit cell with edge length $405 \,pm$. If its density is $2.7 \times 10^{3} \,kg \,m^{-3}$, what is the nature of the cubic unit cell?

Similar Questions

The correct relationship between unit cell edge length '$a$' and radius of sphere '$r$' for face-centred and body-centred cubic structures respectively are:

Calculate the density of an element having molar mass $27 \ g \ mol^{-1}$ that forms $fcc$ unit cell. $[a^3 \cdot N_A = 38.5 \ cm^3 \ mol^{-1}]$ (in $g \ cm^{-3}$)

An element has a body-centered cubic $(bcc)$ structure with a cell edge of $288 \ pm$. The atomic radius is $...... \ pm$.

Calculate the number of unit cells in $1 \, g$ of $NaCl$ crystal,which crystallizes in an $fcc$ structure. (Molar mass of $NaCl = 58.5 \, g/mol$)

The edge length of the unit cell of a metal $(M_W = 24 \, g \, mol^{-1})$ having a cubic structure is $4.53 \, \mathring{A}$. If the density of the metal is $1.74 \, g \, cm^{-3}$,then the effective number of atoms in the unit cell is :- $(N_A = 6 \times 10^{23} \, mol^{-1})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module